Thales's theorem

Thales of Miletus (early 6th century BC) is traditionally credited with proving the theorem; however, even by the 5th century BC there was nothing extant of Thales' writing, and inventions and ideas were attributed to men of wisdom such as Thales and Pythagoras by later doxographers based on hearsay and speculation. Reference to Thales was made by Proclus (5th century AD), and by Diogenes Laërtius (3rd century AD) documenting Pamphila's (1st century AD) statement that Thales "was the first to inscribe in a circle a right-angle triangle". Thales was claimed to have traveled to Egypt and Babylonia, where he is supposed to have learned about geometry and astronomy and thence brought their knowledge to the Greeks, along the way inventing the concept of geometric proof and proving various geometric theorems. However, there is no direct evidence for any of these claims, and they were most likely invented speculative rationalizations. Modern scholars believe that Greek deductive geometry as found in Euclid's Elements was not developed until the 4th century BC, and any geometric knowledge Thales may have had would have been observational. The theorem appears in Book III of Euclid's Elements () as proposition 31: "In a circle the angle in the semicircle is right, that in a greater segment less than a right angle, and that in a less segment greater than a right angle; further the angle of the greater segment is greater than a right angle, and the angle of the less segment is less than a right angle." Dante Alighieri's Paradiso (canto 13, lines 101–102) refers to Thales's theorem in the course of a speech. == Proof ==

First proof The following facts are used: the sum of the angles in a triangle is equal to 180° and the base angles of an isosceles triangle are equal. Since , and are isosceles triangles, and by the equality of the base angles of an isosceles triangle, and . Let and . The three internal angles of the triangle are , , and . Since the sum of the angles of a triangle is equal to 180°, we have :\begin{align} \alpha+ (\alpha + \beta) + \beta &= 180^\circ \\ 2\alpha + 2\beta &= 180^\circ \\ 2( \alpha + \beta ) &= 180^\circ \\ \therefore \alpha + \beta &= 90^\circ. \end{align} Q.E.D. Second proof The theorem may also be proven using trigonometry: Let , , and . Then is a point on the unit circle . We will show that forms a right angle by proving that and are perpendicular — that is, the product of their slopes is equal to −1. We calculate the slopes for and : :\begin{align} m_{AB} &= \frac{y_B - y_A}{x_B - x_A} = \frac{\sin \theta}{\cos \theta + 1} \\[2pt] m_{BC} &= \frac{y_C - y_B}{x_C - x_B} = \frac{-\sin \theta}{-\cos \theta + 1} \end{align} Then we show that their product equals −1: : \begin{align} &m_{AB} \cdot m_{BC}\\[4pt] = {} & \frac{\sin \theta}{\cos \theta + 1} \cdot \frac{-\sin \theta}{-\cos \theta + 1}\\[4pt] = {} & \frac{-\sin ^2 \theta}{-\cos ^2 \theta +1}\\[4pt] = {} & \frac{-\sin ^2 \theta}{\sin ^2 \theta}\\[4pt] = {} & {-1} \end{align} Note the use of the Pythagorean trigonometric identity \sin^2 \theta + \cos^2 \theta = 1. Third proof Let be a triangle in a circle where is a diameter in that circle. Then construct a new triangle by rotating by 180° over the center of the circle . Since we rotated over 180°, lines and are parallel, likewise for and . It follows that the quadrilateral is a parallelogram. The lines and are likewise parallel and meet in , so is a straight line through , meaning is a diameter of the circle. Since lines and , the diagonals of the parallelogram, are both diameters of the circle and therefore have equal length, the parallelogram must be a rectangle. All angles in a rectangle are right angles. Fourth proof The theorem can be proved using vector algebra. Let's take the vectors \overrightarrow{AB} and \overrightarrow{CB}. These vectors satisfy :\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB}\qquad \qquad \overrightarrow{CB} = \overrightarrow{CO} + \overrightarrow{OB} and their dot product can be expanded as :\overrightarrow{AB}\cdot\overrightarrow{CB} = \left(\overrightarrow{AO} + \overrightarrow{OB}\right)\cdot\left( \overrightarrow{CO} + \overrightarrow{OB}\right) =\overrightarrow{AO}\cdot\overrightarrow{CO}+(\overrightarrow{AO}+\overrightarrow{CO})\cdot \overrightarrow{OB}+\overrightarrow{OB}\cdot\overrightarrow{OB} but :\overrightarrow{AO}=-\overrightarrow{CO}\qquad\qquad \overrightarrow{AO}\cdot\overrightarrow{CO}=-R^2\qquad\qquad \overrightarrow{OB}\cdot\overrightarrow{OB} = R^2 and the dot product vanishes :\overrightarrow{AB}\cdot\overrightarrow{CB} = -R^2+\overrightarrow{0}\cdot \overrightarrow{OB}+R^2=0 and then the vectors \overrightarrow{AB} and \overrightarrow{CB} are orthogonal and the angle ABC is a right angle. ==Converse==

For any triangle, and, in particular, any right triangle, there is exactly one circle containing all three vertices of the triangle. This circle is called the circumcircle of the triangle. Its center is called the circumcenter, which is the intersection point of the perpendicular bisectors of the triangle. The locus of points equidistant from two given points is a straight line that is called the perpendicular bisector of the line segment connecting the points. The perpendicular bisectors of any two sides of a triangle intersect in exactly one point. This point must be equidistant from the vertices of the triangle. One way of formulating Thales's theorem is: if the circumcenter lies on the triangle then the triangle is right, and it is on its hypotenuse. The converse of Thales's theorem is then: the circumcenter of a right triangle lies on its hypotenuse. (Equivalently, a right triangle's hypotenuse is a diameter of its circumcircle.) Proof of the converse using geometry This proof consists of 'completing' the right triangle to form a rectangle and noticing that the center of that rectangle is equidistant from the vertices and so is the center of the circumscribing circle of the original triangle, it utilizes two facts: • adjacent angles in a parallelogram are supplementary (add to 180°) and, • the diagonals of a rectangle are equal and cross each other in their median point. Let there be a right angle , a line parallel to passing by , and a line parallel to passing by . Let be the point of intersection of lines and . (It has not been proven that lies on the circle.) The quadrilateral forms a parallelogram by construction (as opposite sides are parallel). Since in a parallelogram adjacent angles are supplementary (add to 180°) and is a right angle (90°) then angles are also right (90°); consequently is a rectangle and lies on the circle. Let be the point of intersection of the diagonals and . Then the point , by the second fact above, is equidistant from , , and . And so is center of the circumscribing circle, and the hypotenuse of the triangle () is a diameter of the circle. Alternate proof of the converse using geometry Given a right triangle with hypotenuse , construct a circle whose diameter is . Let be the center of . Let be the intersection of and the ray . By Thales's theorem, is right. But then must equal . (If lies inside , would be obtuse, and if lies outside , would be acute.) Proof of the converse using linear algebra This proof utilizes two facts: • two lines form a right angle if and only if the dot product of their directional vectors is zero, and • the square of the length of a vector is given by the dot product of the vector with itself. Let there be a right angle and circle with as a diameter. Let M's center lie on the origin, for easier calculation. Then we know • , because the circle centered at the origin has as diameter, and • , because is a right angle. It follows :\begin{align} 0 &= (A-B) \cdot (B-C) \\ &= (A-B) \cdot (B+A) \\ &= |A|^2 - |B|^2. \\[4pt] \therefore \ |A| &= |B|. \end{align} This means that and are equidistant from the origin, i.e. from the center of . Since lies on , so does , and the circle is therefore the triangle's circumcircle. The above calculations in fact establish that both directions of Thales's theorem are valid in any inner product space. ==Generalizations and related results==

As stated above, Thales's theorem is a special case of the inscribed angle theorem (the proof of which is quite similar to the first proof of Thales's theorem given above): :Given three points , and on a circle with center , the angle is twice as large as the angle . A related result to Thales's theorem is the following: • If is a diameter of a circle, then: :*If is inside the circle, then :*If is on the circle, then :*If is outside the circle, then . ==Applications==

Constructing a tangent to a circle passing through a point Thales's theorem can be used to construct the tangent to a given circle that passes through a given point. In the figure at right, given circle with centre and the point outside , bisect at and draw the circle of radius with centre . is a diameter of this circle, so the triangles connecting OP to the points and where the circles intersect are both right triangles. h=\sqrt{pq} with q=1 Finding the centre of a circle Thales's theorem can also be used to find the centre of a circle using an object with a right angle, such as a set square or rectangular sheet of paper larger than the circle. The angle is placed anywhere on its circumference (figure 1). The intersections of the two sides with the circumference define a diameter (figure 2). Repeating this with a different set of intersections yields another diameter (figure 3). The centre is at the intersection of the diameters. ==See also==